If the distance $'s^{'}$metres traversed by a particle in t seconds is given by $s = t^{3} - 3t^{2},$then the velocity of the particle when the acceleration is zero in$m/s\$is

Question

Dr. Abubacker · Accepted Answer

(c)
We have, $s = t^{3} - 3t^{2}$
On differentiating with respect to $t$, we get
$\frac{ds}{dt} = 3t^{2} - 6t$ …(i)
Again differentiating Eq. (i), we get
$\frac{d^{2}s}{dt^{2}} = 6t - 6 = 0\ \  \Rightarrow \ \ t = 1$
On putting the value of $t = 1$ in Eq. (i), we get
$\frac{ds}{dt} = 3 	imes 1 - 6 	imes 1 = 3 - 6 = - 3\ m/s$

Q. If the distance $'s^{'}$metres traversed by a particle in t seconds is given by $s = t^{3} - 3t^{2},$then the velocity of the particle when the acceleration is zero in$m/s\$is

Explanation

Reviewed by Dr. Abubacker

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